Problem 23 - Thomas Calculus

Question

In Exercises \(13-24,\) find the derivative of \(y\) with respect to the appropriate variable. $$y=\left(x^{2}+1\right) \operatorname{sech}(\ln x)$$ (Hint: Before differentiating, express in terms of exponential and simplify.)

Step-by-Step Solution

Answer

The derivative of \( y \) with respect to \( x \) is 2.

Step 1: Rewrite the hyperbolic secant

The hyperbolic secant function is given by \( \operatorname{sech}(x) = \frac{2}{e^x + e^{-x}} \). Rewrite the expression \( y = (x^2 + 1) \operatorname{sech}(\ln x) \) using this identity.

Step 2: Simplify \( \operatorname{sech}(\ln x) \)

Since \( \operatorname{sech}(\ln x) = \frac{2}{e^{\ln x} + e^{-\ln x}} \) and \( e^{\ln x} = x \) and \( e^{-\ln x} = \frac{1}{x} \), we can simplify \( \operatorname{sech}(\ln x) \) to \( \frac{2}{x + \frac{1}{x}} = \frac{2x}{x^2 + 1} \).

Step 3: Simplify the expression for \(y\)

Substitute the simplified form of \( \operatorname{sech}(\ln x) \) back into the expression for \( y \): \[ y = (x^2 + 1) \cdot \frac{2x}{x^2 + 1} = 2x \].

Step 4: Differentiate with respect to \(x\)

Now, differentiate \( y = 2x \) with respect to \( x \): \[ \frac{dy}{dx} = 2 \].

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but relate to hyperbolas instead of circles. One key function in this group is the hyperbolic secant, denoted as \( \operatorname{sech}(x) \), which is related to the exponential function.

• Just as the secant function in trigonometry is defined as the reciprocal of cosine, the hyperbolic secant is the reciprocal of the hyperbolic cosine, \( \operatorname{sech}(x) = \frac{1}{\cosh(x)} \).
• The hyperbolic cosine itself is defined using exponential functions: \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).

To find the derivative of \( y \) in the given exercise, we first needed to express the hyperbolic secant, \( \operatorname{sech}(\ln x) \), using its definition involving exponential functions. By rewriting \( \operatorname{sech}(x) \) as \( \frac{2}{e^x + e^{-x}} \) and plugging in \( x=\ln(x) \), we simplified the function before proceeding with differentiation.

Chain Rule

The chain rule is a fundamental technique for finding the derivative of composite functions. It allows us to differentiate functions that are made up by applying one function inside another.
In the exercise, \( y \) is expressed in terms of hyperbolic secant and natural logarithm functions, leading to a composition that involves both \( x^2+1 \) and \( \operatorname{sech}(\ln x) \). The chain rule essentially tells us how to differentiate such a compound function efficiently.

• The rule states: if \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} \) is \( f'(g(x)) \cdot g'(x) \).
• This means we first differentiate the outer function with respect to the inner function, and then differentiate the inner function with respect to \( x \).

In our case, simplifying the function was crucial before applying differentiation, because handling the hyperbolic and logarithmic functions directly could be cumbersome. Breaking it down helped us apply the chain rule more easily in the step-by-step solution.

Exponential Functions

Exponential functions are essential in calculus and feature prominently in the definition of hyperbolic functions. They have the form \( e^x \), where \( e \) is approximately 2.71828, a mathematical constant.

• Exponential functions are unique because their rate of change is proportional to their value, leading to the exponential growth or decay.
• This is why they are closely tied to hyperbolic functions and were used extensively in the exercise to rewrite \( \operatorname{sech}(\ln x) \).

In the exercise given, rewriting the expression \( \operatorname{sech}(\ln x) \) relied heavily on the properties of exponentials, such as \( e^{\ln x} = x \) and \( e^{-\ln x} = \frac{1}{x} \). Understanding these properties allowed us to simplify the expression effectively,
facilitating the differentiation process that followed. The exponential relationships simplify complexities, making calculus operations more manageable.